If $x = - 2$ is the root of equation $3 x^{2} + 7 x + p = 0$ find the value of show that the root of equation $x^{2} + K (4 x + K - 1) + p = 0$ are equal.

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Solution

Given $x = - 2$ is root of equation $3 x^{2} + 7 x + p = 0$
$∴ x = - 2$ satisfies the given equation
$3 (- 2)^{2} + 7 (- 2) + P = 0$
$12 + (- 14) + P = 0$
$P = 2$
The given equation becomes $3 x^{2} + 7 x + 2 = 0$
$3 x^{2} + 6 x + x + 2 = 0$
$3 x (x + 2) + 1 (x + 2) = 0$
$(3 x + 1) (x + 2) = 0$
$x = \frac{- 1}{3}$ or $x = - 2$
Now, $x^{2} + K (4 x + k - 1) + P = 0$
roots are equal
$∴ D = 0 D = b^{2} - 4 a c$
$x^{2} + 4 x k + k^{2} - k + 2 = 0$
$a = 1 b = 4 k C = k^{2} - k + 2 = 0$
$(4 k)^{2} - 4 (k^{2} - k + 2) = 0$
$16 k^{2} - 4 k^{2} + 4 k - 8 = 0$
$12 k^{2} + 4 k - 8 = 0$
$12 k^{2} + 12 k - 8 k - 8 = 0$
$2 k (k + 1) - 8 (k + 1) = 0$
$k = \frac{8}{12} = \frac{2}{3} k = - 1$ .

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If x= -2 is a root of equation 3x ²+ 7x + p = 0, then find the value of k so that the roots of the equation x²+ k(4x + k - 1) + p=0 are equal.

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x^{2} - x + k = 0

, find the value of p so that the roots of the equation

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[CBSE 2015]

Q. Find the values of

k

for which the given equation has real and equal roots

x^{2} + k (4 x + k - 1) + 2 = 0

Q. If

x^{2} + p x + q = 0

and

x^{2} + q x + p = 0, (p \neq q)

have a common root, show that

1 + p + q = 0

; show that their other roots are the roots of the equation

x^{2} + x + p q = 0

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If x=−2 is the root of equation 3x2+7x+p=0 find the value of show that the root of equation x2+K(4x+K−1)+p=0 are equal.

If $x = - 2$ is the root of equation $3 x^{2} + 7 x + p = 0$ find the value of show that the root of equation $x^{2} + K (4 x + K - 1) + p = 0$ are equal.